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Forces Problem

  1. Oct 7, 2009 #1
    1. The problem statement, all variables and given/known data
    They're two boxes on a ramp. Box 1 (on the bottom) is 8.60kg and has a coefficient of kinetic friction of .35 with an incline plane. Box 2 (on top of Box 1) is 1.3kg and has a coefficient of kinetic friction of .45 between the boxes. The two boxes are connected through a string with a pulley at the end of it. The incline of the ramp is 38 degrees. What is the acceleration of box 1 and box 2?


    2. Relevant equations
    F=ukFn
    F=ma


    3. The attempt at a solution
    Well this is wrong, but I tried to find the frictional force first, so Ffr=.35(66.41N) = 23.24. I found the normal force by doing Fn=mgcos38. Then I just put that force for f=ma and found a. Then I did the same thing for box 2. But, my teacher said their accelerations would be the same. So I am a little bit confused on how to do this, I can't figure it out. Any help would be greatly appreciated. Thank you
     
  2. jcsd
  3. Oct 7, 2009 #2

    Doc Al

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    Staff: Mentor

    Careful: That would be the normal force (between box 1 and the surface) if it wasn't for box 2 on top of it.
    You need to consider all the forces on each box in order to apply ΣF=ma.
    They are connected by a string.

    Hint: Identify all the forces on box 1 and on box 2. (Draw free body diagrams for each.) Then set up force equations for each using Newton's 2nd law.
     
  4. Oct 7, 2009 #3
    Do you have to find the tension from the string to solve it? I am not sure how to find the tension if so. I am still a little confused on how to solve the problem.
     
  5. Oct 7, 2009 #4

    Doc Al

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    The string tension is one of the forces acting on each box. Just call it "T" and set up your equations. The tension and acceleration are the two variables that you can solve for.
     
  6. Oct 7, 2009 #5
    OK thanks, but what is the equation for tension?
     
  7. Oct 7, 2009 #6

    Doc Al

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    Tension is just a force. The equation you need is just Newton's 2nd law.
     
  8. Oct 7, 2009 #7
    Is this how you would do it?

    so M=box1, m=box 2

    a=Mgsin38-.35Mg38-t/M

    a=mgsin38-.45mgcos38-t/m

    solve for T, and get 2.06 N

    then plug it back in the equation and get 4.8 m/s^2?

    can you please tell me if this is right or not. or can you please help me get started if it's wrong!
     
  9. Oct 8, 2009 #8

    Doc Al

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    (1) Don't write t/M. Let all terms on the right be force terms.
    (2) When computing the normal force, use the total weight pressing down on the surface--not just the weight of M.

    Write it like:
    Ma = gravity - friction - T
    How can you solve for T, since a is unknown?

    Correct this equation for block 1 then write a similar equation for block 2. Solve them simultaneously to get T and a.
     
  10. Oct 8, 2009 #9
    Ok that makes sense. So is this how you do it?

    BOX 2:
    ∑Fy=Fnb-m2gcos38-m2ay=0
    Fnb=m2gcos38=10.04 N
    ∑Fx=T-Fb-m2gsin38=m2ax
    (Fb=ukfn)
    T=m2ax + Fb + m2gsin38
    m2ax + ubFnb+m2gsin38
    m2ax + (.45)(10.04)+(1.3)(9.8)sin38
    T=m2ax + 12.36 N

    BOX 1:
    ∑Fy=Fna-m1gcos38-Fnb = m1ay=0
    Fna=m1gcos38 + Fnb
    (8.6)(9.8)cos38 + 10.04 = 76.45
    ∑Fx = m1gsin38 - T - Fa - Fb = m1ax
    = m1gsin38 - m2ax + 12.36N - uaFna - ubFnb = m1ax
    m1ax + m2ax = mgsin38 + 12.36 N - uaFna - ubFnb
    ax=m1gsin38+12.36N-uaFna-ubFnb/m1 + m2
    ax=(8.6)(9.8)sin38+12.36N - (.35)(76.45)-(.45)(10.04)/8.6 + 1.3

    ax = .834 m/s^2

    Is this right? gawd i worked so hard on it :)
     
  11. Oct 8, 2009 #10

    Doc Al

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    Staff: Mentor

    Good.

    Good.
    The sign of the 12.36N term is incorrect in these equations, but I think it's just a typo. (Use parentheses to make your work easier to follow.) The acceleration looks good!

    You bet it's right. Great job! :approve:
     
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